What is the solution for 2y^{''}-7y^'+3y=4x^2 ?

The solution for 2y^{''}-7y^'+3y=4x^2 is y=c_{1}e^{3x}+c_{2}e^{x/2}+(4x^2)/3+(56x)/9+344/27

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2y^{''}-7y^'+3y=4x^2

2 y^{^{''}} - 7 y^{^{'}} + 3 y = 4 x^{2}

y = c_{1} e^{3 x} + c_{2} e^{\frac{x}{2}} + \frac{4 x ^{2}}{3} + \frac{56 x}{9} + \frac{344}{27}

Solution steps

2 y^{''} (x) - 7 y^{'} (x) + 3 y = 4 x^{2}

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{\frac{x}{2}} + \frac{4 x ^{2}}{3} + \frac{56 x}{9} + \frac{344}{27}

y = c_{1} e^{3 x} + c_{2} e^{\frac{x}{2}} + \frac{4 x ^{2}}{3} + \frac{56 x}{9} + \frac{344}{27}

y^{^{''}} + 3 y^{^{'}} + 2 y = (1 + e^{2 x})^{- 1}

y^{^{''}} + 4 y = 4 t^{2}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 4 y^{^{'}} + 5 y = 35 e^{(- 4 x)}

y^{^{''}} - 2 y^{^{'}} + y = \frac{( e ^{x} )}{( x ^{2} + 4 )}

y^{^{''}} - y = 12 t, y (0) = - 12 t

What is the solution for 2y^{''}-7y^'+3y=4x^2 ?
The solution for 2y^{''}-7y^'+3y=4x^2 is y=c_{1}e^{3x}+c_{2}e^{x/2}+(4x^2)/3+(56x)/9+344/27